Okay so this question reminded me of one my brother asked me a while back about the hit day-time novelty-worn-off-now snoozathon Deal or no deal.

For the uninitiated:

In playing deal or no deal, the player is presented with one of 22 boxes (randomly selected) each containing different sums of money, he then asks in turn for each of the 21 remaining boxes to be opened, occasionally receiving an offer (from a wholly unconvincing 'banker' figure) for the mystery amount in his box.

If he rejects all of the offers along the way, the player is allowed to work his way through several (for some unfathomable reason, emotionally charged) box openings until there remain only two unopened boxes: one of which is his own, the other not. He is then given a choice to stick or switch (take the contents of his own box or the other), something he then agonises pointlessly over for the next 10 minutes.

Monty hall

[If you have not seen the monty hall 'paradox' check out this wikipedia link and prepare to be baffled, then enlightened, then disappointed that the whole thing is so trivial. After which feel free to read on.]

There is a certain similarity, you will agree, between the situation a deal or no deal player finds himself in having rejected all offers and the dilemma of Monty's contestant in the classic problem: several 'bad choices' have been eliminated and he is left with a choice between a better and worse choice with no way of knowing between them.

So???

Question: The solution to the monty hall problem is that it is, in fact, better to switch- does the same apply here? Does this depend upon the money in the boxes? Should every player opt for 'switch', cutting the 10 minutes of agonising away???

8 Answers
8

Then no, the Monty Hall solution doesn't apply. The whole point is that the door isn't randomly opened, it's always a goat.

An easy way of seeing this is imagining there are 100 doors, with 99 goats. If, after you pick a door, the host always opens 98 doors of goats, then switching is very intuitively favorable. However, if he had just opened 98 doors at random, then most of the time (98 out of 100) he would open the door with the car behind it; and even on the rare occasions he didn't, you still wouldn't be any better off switching than staying.

See also this answer, in which I try to intuitively explain probability fallacies.

Hmm... any chance you could expand on the last couple of phrases? The rest is nice, but I'm not sure it applies- there is always, comparatively speaking, a car and a goat in the final choice pair.
–
Tom BoardmanJul 28 '10 at 1:57

2

@Tom: Yes of course. But in the Monty Hall problem, switching helps only because the host always opens a door with a goat: he has some extra information which he's using, and the door he opens gives you some information. If he merely opened a door at random and it happened to contain a goat, then there's no incentive to switch; both your current choice and the other one are equally likely to have the car.
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ShreevatsaRJul 28 '10 at 4:06

-Assume you will definately swap.
If you've picked 250k, which is 1/22 chance, you lose.
If you've not picked 250k, which is 21/22 chance, you win.
Every 22 times you play, you would expect to pick 250k once and expect to have 250k in the last 2 boxes twice (22 x 2 = 44).
Of the two times that 250k is in the last 2 boxes, 1 time you will swap (and lose), 1 time you will swap (and win).

-Assume you will definately not swap.
If you've picked 250k, which is 1/22 chance, you win.
If you've not picked 250k, which is 21/22 chance, you lose.
Every 22 times you play, you would expect to pick 250k once and expect to have 250k in the last 2 boxes twice (22 x 2 = 44).
Of the two times that 250k is in the last 2 boxes, 1 time you will not swap (and win), 1 time you will not swap (and lose).

If you swap (or not), half the time you will win, half the time you will lose.
With the monty hall problem, 2/3 or the time if you swap you will win, 1/3 of the time you will lose.
The difference is that in EVERY game of the MH problem, a losing box is revealed and the player reaches the final stage, even though there are more than 2 boxes to start with.
In deal or no deal, 250k would only be at the final stage 2/22 or 1/13 times.

The key difference between the Monty Hall problem and the one you're describing is:
a) the host (in the case of Monty Hall) makes an unsuccessfull choice for the contestant thus altering the probabilities.
b) the host can't provide that information when there is only a binomial distribution (2 choices remaining). If he did your probability would go from .5 to 1. With Monty Hall your probability goes from 1/3 to 2/3 because you have 2 in 3 chances of NOT choosing the goat.

I don't think it matters how you get to the final 2 figures, the question is out of a pool of 20 boxes, or however many it is, what are the chances that I picked the 250k box. The odds are 1 in 20, and that doesn't change. If it's 1 in 20 that I hold the main prize, and it's certainly in 1 of the boxes, then the other box has to have a 19/20 chance of being the top prize (as the chance of it being in one of them is 1/1).

Think of it like this - if it were a 50/50 chance at 2 boxes, you would need to be able to guarantee selection of that box 50% of the time out of 20 boxes - impossible. In 19 out of 20 scenarios, either me picking 1, 2, 3 etc, I only won the 250k prize if I swap, random or with host knowledge is irrelevant.

Something else occurred to me as well - The object of the game is not necessarily to win the 250k, but the highest amount possible. If after the game is played out, I am left with 1p and £2000, I can assume that the original pool I was selecting contains everything between those 2 figures and none of the greater amounts, and adjust the odds accordingly. If there is a 1p, a 50p, £1, £5, £10, £50, £100, £500, £1000 and £2000, and I know with 2 boxes at the end of the game that a 1p and a £2000 are left, then it must still hold that out of those 10 boxes, 9 times I will pick something other than the £2000 box. The fact that I picked the 1p isn't important – it could have been any amount that isn't equal to and below the £2000. In 9 out of 10 cases the £2k is in the other box and I only win the LARGEST AMOUNT I CAN if I swap (given the choice).

Note that the two boxes were selected by me (my own box at the beginning and the remaining box as the only box I did not want to open before). Your argument would imply that in 9 out of 10 cases, the 1p is in the other box. This is not the case. At any moment, the amounts that have not been revealed yet are equidistributed among the boxes that have not been opened yet.
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Hagen von EitzenFeb 11 '13 at 9:57

The probability isn't higher if you switch. In the Monty Hall problem, it's different. Here are the possibilities, if you choose door 1 with Monty Hall:

GOAT goat car
GOAT car goat
Car goat goat

If you choose a goat initially (2/3), Monty will show the other goat. 2/3 times it benefits you to change your door. If you choose the car first (1/3) chance, and you change your door, you will lose the car.

In Deal or No Deal, you open all the cases but the last two, the one you chose (1/25 or however many total cases there are) and the last one you didn't open. Let's say, in the example, they are \$1 and \$1000000.

If you chose \$1 initially and switch, you will win
If you chose \$1000000 initially and switch, you lose.

There is a 50/50 chance you will win a higher value as the cases are independent of each other.

You can really only benefit in beating the banker. If you open everything but \$10000000 and \$1, the banker will give you a good offer, as the million is still there. If you consider this, and your goal is not to get the \$1000000, but the highest amount, then the sample space looks more like this:

The bankers offer, if you consider it, makes it more likely to win, or lose, depending on how you look at it. Once that possibility is eliminated, either by taking the money or not, you are down to a 50/50 chance. It really doesn't make a difference if you switch, because no new information has been given after that last round. Opening the cases can only improve the offer if you're planning on going with the banker, otherwise, the only thing that matters is that last moment. It's like flipping a coin. You can switch your call from heads to tails, but that doesn't change the probability because they are independent of each other.

The way I see it, if your goal is only to get the $1,000,000 prize, then the randomized case opening makes switching useless. However if you modify your goal to getting the larger of the two remaining cases regardless of what the prize amounts are, then Deal or No Deal becomes exactly like the Monty Hall problem and you will want to switch. The randomized revealing of cases is negated by the fact that all the prizes are worth different amounts. When you're down to choosing between your case or switching to the last unopened case, one of the cases is always going to be worth more.

Lets say you're at the final three cases, inside of which are 1 dollar, 5,000 dollars, and 1,000,000 dollars. You randomly choose one of the two cases and OH NO! It was the million dollar case. Now you are playing to win the 5,000 dollars. Forget the million was ever a possibility. As far as you are concerned now, 5,000 dollars is the ultimate prize and every other case was 1 dollar even if they actually said 25,000 or 500,000 dollars. So now, mentally, you have a case that could have 1 dollar or 5,000 dollars and all but one of the other 25 cases have had their contents revealed to be 1 dollar. This makes it effectively the same as the Monty Hall problem exchanging goats and cars for dollar amounts and 3 doors for 26 cases.

I notice that when claiming DoND is not the same as Monty Hall, many people focus of Monty's knowledge of the doors' contents as making the difference. As far as I can tell the only thing Monty Hall is doing is ensuring that there is a value difference between your chosen door and the remaining door. In DoND there is a guaranteed value difference between the cases so the host's knowledge becomes useless. In both games, when you choose your door/case, you have created two groups, group A and group B. Group A is your chosen door/case and Group B is all the others. In the Monty Hall problem, you have a 1/3 chance of the car being in group A and 2/3 of it being in Group B. So when Monty asks you if you want to switch what he's really asking is, "What is more likely, that the car is group A or Group B?" Switching doors means you are saying, "I think the car is more likely to be in Group B." The same thing with DoND. When Howie asks if you want to switch cases he is really asking, "Is it more likely that the 5 grand is in group A or group B?" When you chose a case at random there was only a 1/26 chance it was the 5 grand so that means that there is a 25/26 chance it was in group B. Just like in the Monty Hall Problem, revealing the other cases (either intentionally or randomly) doesn't change the fact that it is much more likely the $5,000 is in group B.

Sometimes I have difficulty properly explaining what I mean so I hope that was all clear and understandable.

In my opinion swapping boxes matters, but the difference is really small one. I am trying to solve the problem this way: when in play only 2 boxes left, 1 million $ not revealed,why it is still unopened?

on 1/22 propability its unopened, because I got lucky and picked it at the beginning: 4,54%

but it also may be unopened, because I opened every other box, but 1 million dollars Whats the propability of opening 20 boxes out of 21, without touching 1 million. the propability goes like: (1-1/21)(1-1/20)...(1-1/2) = 4,76%

So there is 4,54 % chance that You made a lucky pick and 4,76 % chance that You have avoided picking jackpot.